a/ \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

\(\Leftrightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Leftrightarrow m=4\left(tm\right)\)

b/ \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\Leftrightarrow n=10\)

\(\Leftrightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)

a)(1/3)^m=(1/3)^4

b)(3/5)^n=(3/5)^10

c)(-0,25)^p=(-0,25)^4

a)\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

 \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow m=4\)

vay \(m=4\)

b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)

\(\Rightarrow n=10\)

vay \(n=10\)

c) \(\left(-0,25\right)^p=\frac{1}{256}\)

\(\left(\frac{-1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)

\(\Rightarrow p=4\)

voi mu duong thi \(\left(\frac{-1}{4}\right)^4=\left(\frac{1}{4}\right)^4\)

vay \(p=4\)

a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)

\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)

\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow n=4\)

Vậy n = 4

b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)

\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)

\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)

\(\Rightarrow n=3\)

Vậy n = 3

bài 1) Đặt \(B=\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\)

Ta có: \(A=B.\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}\)

\(B.\frac{p}{m-n}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{p}{m-n}=\frac{m-n}{p}.\frac{p}{m-n}+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}\)

\(=1+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}=1+\frac{p}{m-n}.\left(\frac{n-p}{m}+\frac{p-m}{n}\right)\)

\(=1+\frac{p}{m-n}.\left[\frac{\left(n-p\right).n}{mn}+\frac{\left(p-m\right).m}{mn}\right]=1+\frac{p}{m-n}.\frac{n^2-np+pm-m^2}{mn}\)

\(=1+\frac{p}{m-n}.\frac{\left(m-n\right).\left(p-m-n\right)}{mn}=1+\frac{p.\left(m-n\right).\left(p-m-n\right)}{\left(m-n\right).mn}=1+\frac{p.\left(p-m-n\right)}{mn}\)

\(=1+\frac{p^2-pm-pn}{mn}=1+\frac{p^2-p.\left(m+n\right)}{mn}\)

Vì m+n+p=0=>m+n=-p

\(=>B.\frac{p}{m-n}=1+\frac{p^2-p.\left(-p\right)}{mn}=1+\frac{2p^2}{mn}=1+\frac{2p^3}{mnp}\left(1\right)\)

\(B.\frac{m}{n-p}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{m}{n-p}=\frac{m-n}{p}.\frac{m}{n-p}+\frac{n-p}{m}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}\)

\(=1+\frac{m-n}{p}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}=1+\frac{m}{n-p}.\left(\frac{m-n}{p}+\frac{p-m}{n}\right)\)

\(=1+\frac{m}{n-p}.\left[\frac{\left(m-n\right).n}{np}+\frac{\left(p-m\right).p}{np}\right]=1+\frac{m}{n-p}.\frac{mn-n^2+p^2-mp}{np}\)

\(=1+\frac{m}{n-p}.\frac{\left(n-p\right).\left(m-n-p\right)}{np}=1+\frac{m.\left(n-p\right).\left(m-n-p\right)}{\left(n-p\right).np}=1+\frac{m.\left(m-n-p\right)}{np}\)

\(=1+\frac{m^2-mn-mp}{np}=1+\frac{m^2-m\left(n+p\right)}{np}=1+\frac{m^2-m.\left(-m\right)}{np}=1+\frac{2m^2}{np}=1+\frac{2m^3}{mnp}\left(2\right)\) (vì m+n+p=0=>n+p=-m)

\(B.\frac{n}{p-m}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{n}{p-m}=\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}+\frac{p-m}{n}.\frac{n}{p-m}\)

\(=1+\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}=1+\frac{n}{p-m}.\left(\frac{m-n}{p}+\frac{n-p}{m}\right)\)

\(=1+\frac{n}{p-m}.\left[\frac{\left(m-n\right).m}{pm}+\frac{\left(n-p\right).p}{pm}\right]=1+\frac{n}{p-m}.\frac{m^2-mn+np-p^2}{pm}\)

\(=1+\frac{n}{p-m}.\frac{\left(p-m\right).\left(n-p-m\right)}{pm}=1+\frac{n.\left(p-m\right).\left(n-p-m\right)}{\left(p-m\right).pm}=1+\frac{n.\left(n-p-m\right)}{pm}\)

\(=1+\frac{n^2-np-mn}{pm}=1+\frac{n^2-n\left(p+m\right)}{pm}=1+\frac{n^2-n.\left(-n\right)}{pm}=1+\frac{2n^2}{pm}=1+\frac{2n^3}{mnp}\left(3\right)\) (vì m+n+p=0=>p+m=-n)

Từ (1),(2),(3) suy ra :

\(A=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}=\left(1+\frac{2p^3}{mnp}\right)+\left(1+\frac{2m^3}{mnp}\right)+\left(1+\frac{2n^3}{mnp}\right)\)

\(=3+\frac{2p^3}{mnp}+\frac{2m^3}{mnp}+\frac{2n^3}{mnp}=3+\frac{2.\left(m^3+n^3+p^3\right)}{mnp}\)

*Tới đây để tính được m³+n³+p³,ta cần CM được bài toán phụ sau:

Đề: Cho m+n+p=0.CMR: \(m^3+n^3+p^3=3mnp\)

Từ m+n+p=0=>m+n=-p

Ta có: \(m^3+n^3+p^3=\left(m+n\right)^3-3m^2n-3mn^2+p^3=-p^3-3mn\left(m+n\right)+p^3\)

\(=-3mn\left(m+n\right)=-3mn.\left(-p\right)=3mnp\)

Vậy ta đã CM được bài toán phụ

*Trở lại bài toán chính: \(A=3+\frac{2.3mnp}{mnp}=3+\frac{6mnp}{mnp}=3+6=9\)

Vậy A=9

bài 2)

a)Nhận thấy các thừa số của A đều có dạng tổng quát sau:

\(n^3+1=n^3+1^3=\left(n+1\right)\left(n^2-n+1\right)=\left(n+1\right).\left(n^2-n+\frac{1}{4}+\frac{3}{4}\right)\)

\(=\left(n+1\right).\left(n^2-2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n+1\right).\left[\left(n-\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]\)

\(n^3-1=n^3-1^3=\left(n-1\right)\left(n^2+n+1\right)=\left(n-1\right).\left(n^2+n+\frac{1}{4}+\frac{3}{4}\right)\)

\(=\left(n-1\right).\left(n^2+2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n-1\right).\left[\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]\)

suy ra \(\frac{n^3+1}{n^3-1}=\frac{\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]}{\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]}\)

Do đó: \(\frac{2^3+1}{2^3-1}=\frac{\left(2+1\right).\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right).\left[\left(2+0,5\right)^2+0,75\right]}=\frac{3.\left(1,5^2+0,75\right)}{1.\left(2,5^2+0,75\right)}\)

\(\frac{3^3+1}{3^3-1}=\frac{\left(3+1\right).\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right).\left[\left(3+0,5\right)^2+0,75\right]}=\frac{4.\left(2,5^2+0,75\right)}{2.\left(3,5^2+0,75\right)}\)

...........................

\(\frac{10^3+1}{10^3-1}=\frac{\left(10+1\right).\left[\left(10-0,5\right)^2+0,75\right]}{\left(10-1\right).\left[\left(10+0,5\right)^2+0,75\right]}=\frac{11.\left(9,5^2+0,75\right)}{9.\left(10,5^2+0,75\right)}\)

\(=>A=\frac{3\left(1,5^2+0,75\right).4\left(2,5^2+0,75\right)........11.\left(9,5^2+0,75\right)}{1\left(2,5^2+0,75\right).2.\left(3,5^2+0,75\right)........9\left(10,5^2+0,75\right)}=\frac{3.4........11}{1.2......9}.\frac{1,5^2+0,75}{10,5^2+0,75}\)

\(=\frac{10.11}{2}.\frac{1}{37}=\frac{2036}{37}\)

Vậy A=2036/37

b) có thể ở chỗ 1+1/4 bn nhầm,phải là \(1^4+\frac{1}{4}\) ,mà chắc cũng chẳng sao,vì 1⁴=1 mà

Nhận thấy các thừa số của B có dạng tổng quát:

\(n^4+\frac{1}{4}=n^4+n^2+\frac{1}{4}-n^2=\left(n^2\right)^2+2.n^2.\frac{1}{2}+\frac{1}{4}-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)

\(=\left(n^2+\frac{1}{2}-n\right)\left(n^2+\frac{1}{2}+n\right)\)

\(B=\frac{\left(1^2+\frac{1}{2}-1\right).\left(1^2+\frac{1}{2}+1\right).\left(3^2+\frac{1}{2}+3\right).\left(3^2+\frac{1}{2}-3\right)..........\left(9^2+\frac{1}{2}-9\right).\left(9^2+\frac{1}{2}+9\right)}{\left(2^2+\frac{1}{2}-2\right).\left(2^2+\frac{1}{2}+2\right).\left(4^2+\frac{1}{2}-4\right).\left(4^2+\frac{1}{2}+4\right)......\left(10^2+\frac{1}{2}-10\right).\left(10^2+\frac{1}{2}+10\right)}\)

Mặt khác,ta cũng có: \(\left(a+1\right)^2-\left(a+1\right)+\frac{1}{2}=a^2+2a+1-a-1+\frac{1}{2}=a^2+a+\frac{1}{2}\)

Suy ra \(B=\frac{1^2+\frac{1}{2}-1}{10^2+\frac{1}{2}+10}=\frac{1}{221}\)

Vậy B=1/221

a) Câu này thiếu đề nhé bạn.

b) \(\frac{25}{5^n}=5\)

\(\Rightarrow5^n=25:5\)

\(\Rightarrow5^n=5\)

\(\Rightarrow5^n=5^1\)

\(\Rightarrow n=1\)

Vậy \(n=1.\)

c) \(\frac{81}{\left(-3\right)^n}=-243\)

\(\Rightarrow\left(-3\right)^n=81:\left(-243\right)\)

\(\Rightarrow\left(-3\right)^n=-\frac{1}{3}\)

\(\Rightarrow\left(-3\right)^n=\left(-3\right)^{-1}\)

\(\Rightarrow n=-1\)

Vậy \(n=-1.\)

e) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)

\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow n=4\)

Vậy \(n=4.\)

f) \(\left(-\frac{3}{4}\right)^n=\frac{81}{256}\)

\(\Rightarrow\left(-\frac{3}{4}\right)^n=\left(-\frac{3}{4}\right)^4\)

\(\Rightarrow n=4\)

Vậy \(n=4.\)

Chúc bạn học tốt!

d) \(\frac{1}{2}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\left(\frac{1}{2}+4\right)=288\)

\(\Rightarrow2^n.\frac{9}{2}=288\)

\(\Rightarrow2^n=288:\frac{9}{2}\)

\(\Rightarrow2^n=64\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

Vậy \(n=6.\)

g) \(-\frac{512}{343}=\left(-\frac{8}{7}\right)^n\)

\(\Rightarrow\left(-\frac{8}{7}\right)^n=\left(-\frac{8}{7}\right)^3\)

\(\Rightarrow n=3\)

Vậy \(n=3.\)

h) \(5^{-1}.25^n=125\)

\(\Rightarrow5^{-1}.5^{2n}=5^3\)

\(\Rightarrow5^{-1+2n}=5^3\)

\(\Rightarrow-1+2n=3\)

\(\Rightarrow2n=3+1\)

\(\Rightarrow2n=4\)

\(\Rightarrow n=4:2\)

\(\Rightarrow n=2\)

Vậy \(n=2.\)

k) \(3^{-1}.3^n+6.3^{n-1}=7.3^6\)

\(\Rightarrow3^{n-1}+6.3^{n-1}=7.3^6\)

\(\Rightarrow3^{n-1}.\left(1+6\right)=7.3^6\)

\(\Rightarrow3^{n-1}.7=7.3^6\)

\(\Rightarrow n-1=6\)

\(\Rightarrow n=6+1\)

\(\Rightarrow n=7\)

Vậy \(n=7.\)

Chúc bạn học tốt!

b)\(\frac{25}{5^n}\)=5

\(5^n\)=25/5

\(5^n\)=\(5^1\)

⇒ n = 1

c) \(\frac{81}{\left(-3\right)^n}\)=-243

\(\left(-3\right)^n\)=81/-243

\(\left(-3\right)^n\)=\(\frac{-1}{3}\)

\(\left(-3\right)^n\)=\(\left(-3\right)^{-1}\)

⇒n=-1

C = \(25.\left(\frac{-1}{3}\right)^3\) \(+\frac{1}{5}\) \(-2.\left(\frac{-1}{2}\right)^2\) \(-\frac{1}{2}\)

C = \(25.\left(\frac{-1}{27}\right)+\frac{1}{5}\) \(-2.\frac{1}{4}\) \(-\frac{1}{2}\)

C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-\frac{1}{2}\) \(-\frac{1}{2}\)

C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-1\)

C = \(\frac{-125}{135}\) \(+\frac{27}{135}\) \(-\frac{135}{135}\)

C = \(\frac{-233}{135}\)

D =  \(-8.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)

D = \(-8.\frac{1}{2}\) \(.\frac{12}{13}\)

D = \(-4.\frac{12}{13}\)

D = \(\frac{-48}{13}\)

E = \(5\sqrt{16}\) \(-4\sqrt{9}\) \(+\sqrt{25}\) \(-0,3\sqrt{400}\)

E = \(5.4-4.3+5-0,3.20\)

E = \(20-12+5-6\)

E = \(8+\left(-1\right)\)

E = \(7\)

F = \(\left(\frac{-3}{2}\right)\) \(+\left|\frac{-5}{6}\right|\) \(-1\frac{1}{2}\) \(:6\)

F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{3}{2}\) \(.\frac{1}{6}\)

F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{1}{4}\) 

F = \(\left(\frac{-18}{12}\right)\) \(+\frac{10}{12}\) \(-\frac{3}{12}\)

F = \(\frac{-11}{12}\)

 Chúc cậu hk tốt ~ 

a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)

\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)

\(=10:\left(\frac{-2}{3}\right)\)

\(=-15\)

b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\frac{-8}{27}+\frac{1}{10}\)

\(=\frac{-8}{3}+\frac{1}{10}\)

\(=\frac{-77}{30}\)

c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)

\(=\frac{-1}{3}\)

\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)

\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)

\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)

\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)

\(=-15\)

\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)

\(=-\frac{8}{3}+\frac{1}{10}\)

\(=-\frac{77}{30}\)

\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)

\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)

\(=\frac{1}{3}\)